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38=x^2+2x+3
We move all terms to the left:
38-(x^2+2x+3)=0
We get rid of parentheses
-x^2-2x-3+38=0
We add all the numbers together, and all the variables
-1x^2-2x+35=0
a = -1; b = -2; c = +35;
Δ = b2-4ac
Δ = -22-4·(-1)·35
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-12}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+12}{2*-1}=\frac{14}{-2} =-7 $
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